Python3, 259 bytes

lambda X:[(m,n,r)for m in range(1,X+1)for n in range(m+1,X+1)for r in range(1,X+1)if F(n,m,r)and m+n<2*r]
def F(n,m,r):
 T=r**2-((n+m)/2)**2
 if T<0:return 0
 L=2*T**.5
 U=abs((-L+(L**2+4*m*n)**.5)/2)
 return U and int(U)==U and sorted([U,L+U])!=sorted([n,m])

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Python3, 256 bytes

lambda X:[(m,n,r)for m in range(1,X+1)for n in range(m+1,X+1)for r in range(1,X+1)if F(n,m,r)and m+n<2*r]
def F(n,m,r):
 if(T:=r**2-((n+m)/2)**2)<0:return 0
 L=2*T**.5
 return int(U:=abs((-L+(L**2+4*m*n)**.5)/2))==U and U and sorted([U,L+U])!=sorted([n,m])

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Python3, 257 bytes

lambda:[(m,n,r)for m in range(1,101)for n in range(m+1,101)for r in range(1,101)if F(n,m,r)and m+n<2*r]
def F(n,m,r):
 T=r**2-((n+m)/2)**2
 if T<0:return 0
 L=2*T**.5
 U=abs((-L+(L**2+4*m*n)**.5)/2)
 return U and int(U)==U and sorted([U,L+U])!=sorted([n,m])

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Python3, 259 bytes

lambda X:[(m,n,r)for m in range(1,X+1)for n in range(m+1,X+1)for r in range(1,X+1)if F(n,m,r)and m+n<2*r]
def F(n,m,r):
 T=r**2-((n+m)/2)**2
 if T<0:return 0
 L=2*T**.5
 U=abs((-L+(L**2+4*m*n)**.5)/2)
 return U and int(U)==U and sorted([U,L+U])!=sorted([n,m])

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Python3, 234 bytes

The solution below produces all m, n, r triples specified in the original post, and also includes valid outputs which are not contained in the test cases.

lambda:[(m,n,r)for m in range(1,101)for n in range(m+1,101)for r in range(1,101)if F(n,m,r)and m+n<2*r and n>m]
def F(n,m,r):
 T=r**2-((n+m)/2)**2
 if T<0:return 0
 L=2*T**.5
 F=(L**2+4*m*n)**.5
 return(-L+F)/2 and int(U:=(-L+F)/2)==U

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Python3, 257 bytes

lambda:[(m,n,r)for m in range(1,101)for n in range(m+1,101)for r in range(1,101)if F(n,m,r)and m+n<2*r]
def F(n,m,r):
 T=r**2-((n+m)/2)**2
 if T<0:return 0
 L=2*T**.5
 U=abs((-L+(L**2+4*m*n)**.5)/2)
 return U and int(U)==U and sorted([U,L+U])!=sorted([n,m])

Try it online!